Answer:
Explanation:
a ) It is given that bomb was at rest initially , so , its momentum before the explosion was zero.
b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.
If v be the velocity of the third part along a direction making angle θ
with x axis ,
x component of v = vcosθ
So momentum along x axis after explosion of third part = mv cosθ
= 10 v cosθ
Momentum along x of first part = - 5 x 42 m/s
momentum of second part along x direction =0
total momentum along x direction before explosion = total momentum along x direction after explosion
0 = - 5 x 42 + 10 v cosθ
v cosθ = 21
Similarly
total momentum along y direction before explosion = total momentum along y direction after explosion
0 = - 5 x 38 + 10 v sinθ
v sinθ= 21
squaring and and then adding the above equation
v² cos²θ +v² sin²θ = 21² +19²
v² = 441 + 361
v = 28.31 m/s
Tanθ = 21 / 19
θ = 48°
Answer:
after 2 seconds its velocity is -20 m/s. after 3 seconds its velocity is -30 m/s. after 10 seconds its velocity is -100 m/s.
Explanation:
This is my answer.
Mass is how heavy is it, weight is the size both are the same
The correct answer is
C ). A hypothesis includes an explanation for why two variables affect each other, but a law only describes how they affect each other.
How tall is the building?
u need to divide the height by 1.6 to get your answer
